local mean value property for subharmonic functions . $u$ satisfies  "mean value property locally " on $\Omega$ if for every $x\in \Omega \exists \delta=\delta (x)>0 $   such that   
$u(x) \le \frac {1}{|\mu(B(x,r)|}\int_{\partial B(x,r)} u(y) dS_y$ 
for all $r\le \delta(x)$
Does this imply that if $u\in C^2(\Omega)$ and satisfies mean value property locally in $\Omega$ then $u$ is subharmonic ?? 
Any hints will be nice. 
 A: We first claim that if a $C^0(\bar{\Omega})$ function $u$ satisfies the following local mean value inequality 
$$u(y)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(y)}u  ds \text{ for all } R\leq \delta$$ 
then maximal principle applies, i.e. $u$ attains its maximum at $\partial \Omega$.
Assume our claim now. Let $\tilde{u}$ be the harmonic function defined by the boundary value $u|_{\partial B}$ on the boundary of an arbitrary ball $B\subset\subset \Omega$. Then $\tilde{u}$  satisfies mean value equality in $B$ since it is harmonic in $B$. Then $u-\tilde{u}$ satisfies the above mean value inequality. Then by our claim (maximal principle), we have $\sup_{x\in B}(u-\tilde{u})=\sup_{x\in\partial B}(u-\tilde{u})$. But $u-\tilde{u}=0$ on $\partial B$. Hence we have $u-\tilde{u}\leq 0$. $u\leq \tilde{u}$ in $B$. 
Now we show that $u$ is subharmonic, given an harmonic function $h$  in $B\subset \subset \Omega$, such that $u\leq h$ on $\partial B$. We need to show that $u\leq h$ in B. Since we know $u\leq\tilde{u}$ in $B$. It suffices to show that $\tilde{u}\leq h$ in $B$. To see this one uses the Possion kernel $$\tilde{u}(x)=\int_{\partial B}K(x,y)u(y)ds_y$$
$$h(x)=\int_{\partial B}K(x,y)h(y)ds_y$$
 Since $u(y)\leq h(y)$ on $\partial B$ and $K(x,y)\geq 0$. We have $\tilde{u}(x)\leq h(x)$.  
Now we show our claim that the local mean value inequality
$u(y)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(y)}u  ds $ 
implies maximal principle. 
We use the standard topological argument. Define a set $S:=\{q\in\bar{\Omega}| u(q)=\sup_{x\in\bar{\Omega}}u(x)\}$. If $u$ attains maximum at a point $p\in\Omega$ then $S$ is nonempty since $p\in S$. Note that $S$ is closed since every sequence in $S$ converges to a point in $S$ since $u$ is continuous. Next if  $p\in S\cap\Omega$, then  $u(p)\leq\frac{1}{n\omega_nR^{n-1}}\int_{\partial B_R(p)}u  ds $ and $u(p)\geq u(x)$ implies $u(x)=u(p)$ for all $x\in B_R(p)$ since $u$ is continuous. Hence $B_R(p)\in S$
Hence $S$ is open. Now $S$ is open and closed and nonempty, hence we have $S=\bar{\Omega}$. This proves our claim and finished our proof.
A: No, a function satisfying a mean-value inequality is not necessarily $C^2$ (think of the norm function $\mathbb R^d\supseteq\Omega\ni\vec x\mapsto|\vec x|\in\mathbb R$) but if it satisfies a mean-value equality it is harmonic and therefore $C^\infty$ (in fact, analytic).
Yes, a function's being (sub/super)mean-valued is (sub/super)harmonic, but you need to interpret the Laplacian in a generalized sense. For example, in my example above the Laplacian is a dirac-delta which is a positive distribution (testing against positive test functions yields a positive output).
